Question

The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What is the correct standard deviation?

• A. 3
• B. 4
• C. 2
• D. 5

Answer 1

Answer: (D)

Given, $n=100$, Incorrect mean, $\bar{x}=40$, Incorrect standard deviation, $\sigma=5.1$
Now, $\bar{x}=\frac{\Sigma x_i}{n}$
$\Rightarrow 40=\frac{\Sigma x_i}{100}$
$\Rightarrow \Sigma x_i=4000$
Correct $\Sigma x_i=$ Incorrect $\Sigma x_i-50+40$
$=4000-50+40=3990$
$\therefore$ Correct $\bar{x}=\frac{\text { Correct } \Sigma x_i}{n}=\frac{3990}{100}=39.9$
Also, $\sigma=\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2}$
$\Rightarrow 5.1=\sqrt{\frac{\sum x_i^2}{100}-(40)^2}$
$\Rightarrow(5.1)^2=\frac{\sum x_i^2}{100}-1600$
$\Rightarrow 26.01+1600=\frac{\sum x_i^2}{100}$
$\Rightarrow \sum x_i^2=162601$
Correct $\sum x_i^2=\text { Incorret } \sum x_i^2-(50)^2+(40)^2 \\ =162601-(50)^2+(40)^2=161701$
$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_i^2}{n}-(\text { Correct } \bar{x})^2}=\sqrt{\frac{161701}{100}-(39.9)^2}$
$=\sqrt{1617.01-1592.01}=\sqrt{25}=5$